# Why does M4 turn off and Vout be equal to zero?

From this picture,when$V_{in} = V_{DD}$,I know $M_1$ and $M_3$ will turn on,and $M_2$ will turn off because their $V_{gs}$ are $V_{DD}$ and 0 ,but why will $M_4$ turn off,and $V_B=V_{DD}$, $V_{out}=0$ ?

If i know $V_{out}=0$ and $V_B=V_{DD}$ first,then of course i know $M_4$ will turn off ,but if now i don't know the value of $V_B$ and $V_{out}$,and neither do $M_4$ ,how do i know $M_4$ turn off,and $V_B=V_{DD}$, $V_{out}=0$ ?

• How about recognizing good answers on previously raised questions or do you think you deserve free advice? Feb 2, 2018 at 14:56
• i had did it,but some question is not the right answer Feb 2, 2018 at 14:58
• but some question is not the right answer Hmm, does that mean you already know what the right answer is? Feb 2, 2018 at 15:05
• why M4 and M3 gates are floating ?? Feb 2, 2018 at 15:19
• @TonyStewart.EEsince'75 That's what I thought as well but it is just drawn in a bad way. That node is actually $V_{IN}$. A simple vertical line to the $V_{IN}$ text would have fixed that, like is done at the left side of the circuit. Feb 2, 2018 at 15:22

As the gate of M4 (which is a PMOS) is $V_{DD}$, the only way to make M4 turn on is the take the bulk or drain or source connection of M4 at least $V_t$ above $V_{DD}$. So we would need $V_{DD} + V_t$ to be present somewhere in the circuit.
This can only happen if capacitor $C_B$ is first charged (with $V_A$ = GND and $V_B$ = $V_{DD}$) and then discharged by making $V_A = V_{DD}$ then there would for a short time be $2*V_{DD}$ present at $V_B$. But as $C_B$ discharges through M5, which is on, see Andy's answer, so the $V_{GS}$ of M4 must be zero, hence it is off.